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Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$

My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$

$(a-b)^2\geq0\\a^2+b^2\geq2ab \\\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}$

I do not know how to link them together. Appreciate any tips.

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Run this backwards:
$$\frac{a+b}2-\sqrt{ab}\geq\sqrt{\frac{a^2+b^2}2}-\frac{a+b}2\\ a+b\geq\sqrt{ab}+\sqrt{\frac{a^2+b^2}2}\\ a^2+2ab+b^2\geq ab+\frac{a^2+b^2}2+2\sqrt{ab\frac{a^2+b^2}2}\\ a^2+2ab+b^2\geq4\sqrt{ab\frac{a^2+b^2}2}\\ (a+b)^4\geq 8(a^3b+ab^3)\\ (a-b)^4\geq0$$

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By Bernoulli's Inequality, $$\begin{align} \sqrt{\frac{a^2+b^2}{2}} &=\sqrt{\frac{(a+b)^2+(a-b)^2}{4}}=\frac{a+b}{2}\left(1+\left(\frac{a-b}{a+b}\right)^2\right)^{\frac12} \\ &\leq \frac{a+b}{2}\left(1+\frac{1}{2}\left(\frac{a-b}{a+b}\right)^2\right)\,. \end{align}$$ Thus, $$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}\leq \frac{(a-b)^2}{4(a+b)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{4(a+b)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\left(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^2}{\frac{a+b}{2}}\right)\,.$$ Now, either by the AM-GM Inequality, by the Cauchy-Schwarz Inequality, or by the Power-Mean Inequality, we have $\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^2\leq \frac{a+b}{2}\,.$ That is, $$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}\leq \frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\left(\frac{\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^2}{\frac{a+b}{2}}\right)\leq\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}=\frac{a+b}{2}-\sqrt{ab}\,.$$

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By setting: $$ M_p(a,b)=\lim_{q\to p^+}\left(\frac{a^q+b^q}{2}\right)^{\frac{1}{q}}$$ we have to prove: $$2M_1\geq M_0+M_2, $$ but $M_0^2+M_2^2 = 2M_1^2$, hence the previous line follows from Cauchy-Schwarz inequality:

$$ M_0+M_2 \leq \sqrt{2}\sqrt{M_0^2+M_2^2} = 2M_1.$$

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